Understanding Watt Meters

[Geoff S]

Mike, that leaflet doesn't agree with the motor specs. 911 watts on an 11.1 battery needs an 80 amp current draw to a first approximation. The motor is specified to have a maximum current of only 50 amps. So the maximum power with a 3S pack is 11 x 50 = 550 watts! It's better to use higher voltages (ie a 4S or 5S pack) once you're getting over a need for 500 watts plus. Running a prop that size is likely to burn out the motor or the esc if it isn't rated at a high enough current.

As I said earlier the quoted power of an electric motor shouldn't be taken as the power it will deliver as is the case with ic engines. It depends on using the right battery (ie number of cells or voltage) for the maximum allowed current. I much prefer to use a bigger motor than needed coupled with an esc with a good current overhead and then adjust prop size to provide the power you need whilst keeping an eye on the current draw.

Electric motors are much more versatile than engines but that means you need to 'tune' them for your needs.

Geoff

[Martin Harris - Moderator]

I don't know about the leaflet but my interpretation of the figures on the HK site is that the 910W corresponds very closely with the maximum cell count (5S or nominal 18.5V) at 50 Amps.

Edited By Martin Harris on 24/09/2017 22:11:09

[Nigel R]

Mike

Your 4S figures are quite close to those of my dad's "40 replacement" setup.

 

Geoff

"I haven't run 40 size glow engines for a while but I seem to remember that the usual prop was a 10x6 so start with that. Flat out that should be turning at about 10k rpm."

A Schneurle 40 will produce around 600W at the prop, probably 10k on an 11x6. So an 800W electric setup is a close replacement.

An FP type 40 would make a bit less, I guess 500W?

Going back to an old crossflow 40 type, more like 400W at the prop.

edit: These would all be a standard silencer setup, normal fuel, propped for reasonable revs not screaming, a normal club model setup in other words. I've got the power numbers from the Pe Reivers spreadsheet, as they are all calculated take with a pinch of salt.

Edited By Nigel R on 25/09/2017 11:37:32

[Peter Beeney]

I think I would always be a bit cautious when it comes to directly comparing the output power of the battery with the output power of the motor; or indeed the output power of an i.c. engine, too. To my mind we are measuring two different types of movement here.

I’ve probably got all this wrong again, I usually seem so to do, but as I see it the battery power measurement is electrical whilst the motor power measurement is a mechanical one. They are always going to equal each other in total, but the two different effects are not always what we really want.

Generally speaking,* when a current passes through a conductor it has to overcome the resistance, an action which produces heat, and it at the same time it creates a magnetic field surrounding the conductor. Again, rightly or wrongly, I figure that the magnetic field comes for free; the cost, in our case here anyway, is that it always requires it’s accompanying current and that in turn is causing heat which we don’t want, at least in this case.

The circular motion of our permanent magnet DC brushless motor relies solely on the reaction, or rather a deflecting action, of two magnetic fields, one established by the flow of current. The mechanical movement, or power, of the prop shaft is derived by multiplying the torque by the rpm and here we seem to be measuring the strength of the inter-action between the magnetic fields; to get an accurate measurement of the power we would need a dynamometer to read the torque and rpm to calculate the power; whereas placing a watt meter between the battery and ESC measures the amps and volts to give the watts. Unfortunately this is just measuring the heat generated, rather than the turning motion!

To illustrate this let’s consider two points on the motor’s power curve. At the instant of start up the motor is stationary therefore the current flow will be at a maximum. If we were able to read our meters in slow motion the watt meter would read max amps, therefore max watts input (all heating effect) but although the dynamometer would read max torque there are no revolutions which equals zero power output; but of course this situation doesn’t last very long; the motor very quickly starts to turn, the current flow diminishes and the speed and current flow equalise at whatever values they ultimately reach.
Right at the other end of the curve let’s consider the motor in a stalled state after running. Exactly the same situation, max watt meter reading into the ESC and motor windings but no power indicated by the dynamometer, again because there are no rpm’s. As before we are just seeing the heating effect, this time possibly indicated by the distress signals given by the power train in the form of an arising column of white smoke or worse. In my view these are the two extreme points and somewhere in between there is one point where the motor will run at it’s ‘best case’ compromise speed, where the two measurements are closest together. I suspect, in very general terms, this is most likely to be not too far away from the unloaded speed, but there are many different cases.

For me this is an indicator of the importance of the resistance, or rather lack of it, of the total power train. Looking at a fictitious case of 3 circuits, with 3 different total resistances but the same 10 amps flowing in each, a) 1ohm, b) 0.1ohm and c) 0.01ohm; Then in circuit a) there would be 100 watts dissipated as heat, circuit b) 10W dissipated and circuit c) 1W. All hypothetical of course, to maintain 100 amps three different voltages would have to be applied, from a) 10V, b) 1V and down to c) 0.1V. But the magnetic field strength would be the same in each case so the mechanical power output would not change. I’d judge the quality a power train by the value of it’s (low) resistance and by the same token I’d always try and keep the rpm as high as possible, within the required performance envelope of course.

* When some conductors are ‘super cooled’, that’s frozen down to near absolute zero, they tend to lose their resistance. In this state a vast amount of electricity could flow unimpeded, there is no heat generated by this but a very strong magnetic field could be created. This can be used to an advantage, indeed I believe I’ve read somewhere that the Large Hadron Collider at Cern uses super cooled electro magnets in the particle accelerator. I know nothing about it, though, does it also make it cheap(er)? to run, too? But I reckon one prevailing caveat at least must be the fountain-like eye watering cost of maintaining the low low low temperature…

Just my idea on the subject and of course it does not take into consideration the other principles that enable the motor to function…and no assurances that any of this is correct, either…

PB

[Frank Skilbeck]

ECalc gives you the electric power in and the mechanical power out, in the case of this motor on a 12 x 8 prop, it calculates 680w in (electrical) and 535w out (mechanical), which makes this combination around 79% efficient .

Note the highest mechanical efficiency doesn't always occur when you keep the rpm as high as possible and larger slower turning propellers are much better at converting the mechanical power to thrust.

[Nigel R]

"to get an accurate measurement of the power we would need a dynamometer to read the torque and rpm to calculate the power"

Don't propellers make a reasonable dynamometer?

I think you may be overthinking the rest of it, power is power whatever form it is in. The power out of the battery, about 80% goes toward turning the shaft, the rest is lost as heat in ESC, motor and to a lesser extent the battery and wiring. 80% is a good enough start point, to figure what sort of power you are getting at the prop.

Its reasonable to compare that figure directly to a shaft hp rating on an IC.

Obviously these powers can be achieved with different props at different revs, but that's another discussion.

Edited By Nigel R on 27/09/2017 15:05:59

[Mike Blandford]

A motor has a "kv" value (rpm/volt). What this specifies is the voltage the motor generates when turning. So, if you have a motor with a kv value of 1000 rpm/volt, and it is turning at 10000 rpm, then the motor generates 10 volts (acting like a dynamo).

When the motor has a voltage applied to it to make it turn, the motor still generates this voltage, which is in the opposite sense to the applied voltage. The mechanical output power, in watts, is simply the current flowing multiplied by this generated voltage.

There will also be some mechanical losses. These may be estimated by finding the current taken by the motor when it runs without a load (propellor in our case). The mechanical losses are aproximately the no load current multiplied by the generated voltage.

Here is an example, take a motor with a kv of 1000. Test it with no load (or use the manufacturers sepcification) to find the no load current. Let us assume this is 2 Amps.

Now we run the motor with a load. A wattmeter shows an input voltage of 12V and a current of 30 A giving an input power of 360 Watts. The motor is also measured to by running at 10000 rpm, so is generating a voltage of 10 volts.

Of the 30A, 2A is overcoming the mechanical losses, so 28A is left providing the output power. The output power is therefore 28A multiplied by 10V (generated voltage) or 280 Watts, at an efficiency of about 78% (280/360).

2A (no load current) times 10V (generated voltage) is 20W so this is the power loss due to mechanical losses.

Of the 360W going in to the motor, 280W is useful output power, 20W is mechanical losses leaving 60W as electrical losses. This value may also be calculated by (applied voltage-generated voltage) mutiplied by current = (12-10)*30 = 60.

Most of this is heat in the windings of the motor.

Suppose we load the motor with a larger propellor so it only spins at 9000 rpm. The wattmeter now shows 12V (what a good battery I'm using!) and 45A, so 540 watts in.

The generated voltage is now only 9V, so the output power is 387 watts ((45-2)*9), mechancial losses are 18W (2*9) and the electrical losses are 135W. The efficiency is now around 72%, but notice how much hotter the motor wil become with 135W instead of 60W of heat generated in the windings.

Mike.

[Peter Beeney]

Exactly so, Mike, but I’m still a little sceptical in parts. Multiplying the amps by the generated voltage to get the output power is surely just measuring the heat generated again, not the mechanical turning force? This may well be ok when the input and output power ratings are as close together as they are going to get, but not so accurate when they are far apart. Your 9V example in the last paragraph, as apposed to the previous 10V, would appear to show the way this is developing. You have the ability to do the calculations so you would be well aware of whats what but I might simply look at the 387 watts displayed on my watt meter and think what a nice powerful motor I’ve got; this is exactly the reason why I’m saying I would be a bit cautious when comparing figures. Also the sentence: ‘The motor is also measured to by running at 10,000 rpm,’ does that imply you’ve also clocked the rpm with a tachometer? If so, how often is this action quoted, recommended or even mentioned? I seem to have missed it. As it happens, a tacho is always right on the top tray of my toolbox together with a new CR 2025 cell, it’s bound to go flat just when I need it…I’ve found with a bit of practise this can give me a great deal of information.

Frank, I think I would like to try out the bits and pieces listed in Mike Downs’s OP on the bench for myself. For instance, what (low) cell count pack would I be using to keep the amps at 50 or below on a 12 x 8 prop? Also what are the revs, this is where I start, it will give me a good indication how the model will fly amongst other things. Mike B mentioned his ‘good’ 12 volt battery, (able to sustain 12v under that sort of load!), I use a fully charged 12V car battery for bench testing motors, now that is a nice stable supply! If I had a model that required a larger slower propeller I’d want to obtain an appropriate motor that still allows me to use it at it’s optimum revolutions point.

Nigel, I’m not sure a propeller is any sort of dynamometer at all, but with a bit of trial and error, or as in my case, poke and hope, it will give you a fairish visual indication of how well or not your model will perform. At the end of the day I probably don’t really need to know very much information at all, I can just tinker until I get it right. But a dynamo. will give me the point on the revolutions curve where it’s at it’s most powerful straightaway. Interestingly, I believe that model aero engines were once tested with a propeller. They needed to know the torque so that together with the rpm they could calculate the horsepower. In the first instance they bolted the engine on a torque beam which was variably weighted at one end and pivoted against the opposite torque of the propeller thrust; but there were some anomalies arising due to the prop wash etc. so they found the the dynamometer a better instrument to use. And somewhere in the distant memory banks I’m sure I’ve seen a picture of an early full size aero engine up to the same sort of caper…
I simply said that I would be a bit cautious when directly comparing the output power of the battery with the output power of the motor; and also including i/c. That’s not to say you shouldn’t do it though, in fact I tend to subconsciously compare electric motors with i/c all the time, I suppose because I’ve used glow engines for so long I’ve a pretty good idea how they are going to perform. Again fine if the the electric set up is spot on but again if the watt meter gives an erroneously high reading it might be a tad misleading. However, I’m sure that as folks are using e.p. more and more the increasing knowledge and experience will lead to less confusion.

I actually think the modern DC brushless motor is a real cracking lump of engineering. With it’s design features and some help from components like neodymium magnets and the very essential but equally as impressive Electronic Switching Commutator it’s a real little powerhouse; and I have noted at the patch that some are still running faultlessly after very much use, too, so in some cases at least their longevity would appear not to be questionable either.

More power to your (electric) prop…

PB

[Mike Blandford]

"Multiplying the amps by the generated voltage to get the output power is surely just measuring the heat generated again, not the mechanical turning force?"

The mechanical turning force (torque) is directly proportional to the current flowing.

The generated voltage (dynamo/alternator effect) opposes the current flow. We are forcing the current against this voltage, so there is no heat generated by this, it is exactly the mechanical power. The heat is generated in the windings, and the amount is the difference between the generated voltage and you battery voltage, multiplied by the current.

A "perfect" motor would have zero resistance in the windings (and no mechanical losses!). In this case, if you apply 12V to a motor with a kv of 1000 rpm/volt, the motor will run at 12000 rpm. It will be generating 12V to oppose the applied voltage.

A "real" motor is a "perfect" motor with a resistor in series. In my examples, the motor has a resistance of 0.066667 ohms. With 30A flowing, this resistance drops 2.0 volts, so the "perfect" motor sees only 10V, so spins at 10000 rpm.

Yes, you do need to measure the motor rpm to obtain the generated voltage (called back e.m.f.).

With all motors, in one respect current is your enemy, unfortunately you need current to generate torque!
Losses due to current flowing are proportional to current squared, double the current then you get four times the losses. Where possible, it is better to use a motor with a lower kv and use more cells at a lower current.

Mike.

[Nigel R]

"Where possible, it is better to use a motor with a lower kv and use more cells at a lower current."

Lower kv motors have more winds of thinner wire with greater resistance. You can't win

"Nigel, I’m not sure a propeller is any sort of dynamometer at all"

What does a dynamometer do?

wiki - "a device for measuring force, torque, or power. "

A given prop needs a given amount of power to turn at a given rpm. If you have some well established props, you can read power by reading rpm on those props. This is why I quote Pe Reivers spreadsheet - it gives a pretty fair estimate of power at the prop, given the prop size and brand.

"modern DC brushless motor is a real cracking lump of engineering. With it’s design features and some help from components like neodymium magnets and the very essential but equally as impressive Electronic Switching Commutator"

Couldn't agree more. That said the ESC is the engineering winner for me, by far. It's a near perfect bundle of software, CPU and power FET development that has culminated in these dirt cheap devices having such long service life, great capability and simplicity of use.

Having LiPo batteries along for the ride certainly hasn't slowed down adoption of the technology

[Mike Blandford]

"Lower kv motors have more winds of thinner wire with greater resistance. You can't win".

You potentially win by lower losses, due to lower current, all the other wiring, the ESC and the battery. Also, if you can use a physically larger motor, then you get the extra windings without needing thinner wire. A larger motor may well be only 30 to 40gm heavier, but everything runs that much cooler, and with some aircraft you need the extra weight in the nose anyway!

Mike.